What is Heat and Specific Heat?

Heat transfer, specific heat capacity, and calorimetry

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Heat and Specific Heat is part of the AP Physics 2 course on Study Mondo, specifically in the Thermodynamics section. You can explore the full course for more related topics and practice resources.

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🔥 Heat and Specific Heat

Heat vs. Temperature

Temperature: Measure of average kinetic energy of particles Heat (Q): Energy transferred between objects due to temperature difference

💡 Key Distinction: Temperature is a property; heat is energy in transit.

Heat flows spontaneously from hot to cold until thermal equilibrium is reached.

Units of Heat

Joule (J): SI unit of energy
Calorie (cal): Energy to raise 1 g of water by 1°C
- 1 cal = 4.186 J
Kilocalorie (kcal or Cal): Food "Calorie"
- 1 kcal = 1000 cal = 4186 J

Specific Heat Capacity

Specific heat (c) is the energy required to raise the temperature of 1 kg of a substance by 1°C (or 1 K).

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

Given:

Mass: $m = 2.0$ kg
Initial temp:

Substance	c (J/kg·°C)
Water	4186
Ice	2090
Steam	2010
Aluminum	900
Steel	470
Copper	387
Lead	128

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? (c_water = 4186 J/kg·°C)

Given:

Mass: $m = 2.0$ kg
Initial temp: $T_i = 20°\text{C}$
Final temp: $T_f = 80°\text{C}$
Specific heat: $c = 4186$ J/kg·°C

Find: Heat required $Q$

Solution:

Step 1: Calculate temperature change. $\Delta T = T_f - T_i = 80 - 20 = 60°\text{C}$

Step 2: Apply heat formula. $Q = mc\Delta T$ $Q = (2.0)(4186)(60)$ $Q = 502,320 \text{ J} = 502 \text{ kJ}$

Answer: 502 kJ of heat is required

This is equivalent to about 120 food Calories (kcal).

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{A l} c_{A l} Δ T_{}$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Step 3: Substitute values. $(0.50)(900)(T_f - 100) + (2.0)(4186)(T_f - 20) = 0$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $Δ T_{w} = 24.1 - 20$ ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

❓ Question:

A 0.50 kg piece of aluminum at 100°C is dropped into 2.0 kg of water at 20°C. What is the final equilibrium temperature? (c_Al = 900 J/kg·°C, c_water = 4186 J/kg·°C)

Given:

Aluminum: $m_{Al} = 0.50$ kg, $T_{Al,i} = 100°\text{C}$ , $c_{Al} = 900$ J/kg·°C
Water: $m_w = 2.0$ kg, $T_{w,i} = 20°\text{C}$ , J/kg·°C

Find: Final temperature $T_f$

Solution:

Step 1: Set up energy conservation. $Q_{Al} + Q_w = 0$ $m_{A l} c_{A l} Δ T_{}$

Step 2: Express temperature changes. $m_{Al}c_{Al}(T_f - T_{Al,i}) + m_w c_w(T_f - T_{w,i}) = 0$

Step 3: Substitute values. $(0.50)(900)(T_f - 100) + (2.0)(4186)(T_f - 20) = 0$

Verification:

Aluminum cools: $\Delta T_{Al} = 24.1 - 100 = -75.9°\text{C}$ ✓
Water warms: $Δ T_{w} = 24.1 - 20$ ✓

Answer: Final temperature is 24.1°C

The water barely warms because it has much larger mass and specific heat!

❓ Question:

A 200 g aluminum cup (c = 900 J/(kg·°C)) contains 500 g of water (c = 4186 J/(kg·°C)) at 20°C. A 100 g piece of copper (c = 387 J/(kg·°C)) at 80°C is placed in the water. Find the final equilibrium temperature.

Solution:

Given:

Al cup: m₁ = 0.200 kg, c₁ = 900 J/(kg·°C), T₁ = 20°C
Water: m₂ = 0.500 kg, c₂ = 4186 J/(kg·°C), T₂ = 20°C
Cu: m₃ = 0.100 kg, c₃ = 387 J/(kg·°C), T₃ = 80°C

Energy conservation: Heat lost by copper = Heat gained by water + aluminum

m₃c₃(T₃ - T_f) = m₁c₁(T_f - T₁) + m₂c₂(T_f - T₂)

Since T₁ = T₂ = 20°C: (0.100)(387)(80 - T_f) = (0.200)(900)(T_f - 20) + (0.500)(4186)(T_f - 20)

3096 - 38.7T_f = 180T_f - 3600 + 2093T_f - 41,860 3096 - 38.7T_f = 2273T_f - 45,460 48,556 = 2311.7T_f T_f = 21.0°C

The large mass and high specific heat of water dominates.

❓ Question:

A 0.20 kg piece of unknown metal at 150°C is placed in 0.50 kg of water at 20°C. The final temperature is 25°C. What is the specific heat of the metal? Assume no heat is lost to surroundings. (c_water = 4186 J/kg·°C)

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_{m} c_{m} Δ T_{m} + m_{w} c_{w} Δ T_{}$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Step 3: Solve for $c_m$ . $m_m c_m \Delta T_m = -m_w c_w \Delta T_w$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

❓ Question:

How much heat is required to raise the temperature of 2.0 kg of water from 20°C to 80°C? Use c_water = 4186 J/(kg·°C).

Solution:

Given: m = 2.0 kg, c = 4186 J/(kg·°C), ΔT = 80 - 20 = 60°C

Heat required: Q = mcΔT Q = (2.0)(4186)(60) Q = 5.02 × 10⁵ J or 502 kJ

❓ Question:

Given:

Metal: $m_m = 0.20$ kg, $T_{m,i} = 150°\text{C}$ , $c_m = ?$
Water: $m_w = 0.50$ kg, $T_{w,i} = 20°\text{C}$ , J/kg·°C
Final: $T_f = 25°\text{C}$

Find: Specific heat of metal $c_m$

Solution:

Step 1: Set up energy conservation. $Q_m + Q_w = 0$ $m_{m} c_{m} Δ T_{m} + m_{w} c_{w} Δ T_{}$

Step 2: Calculate temperature changes. $\Delta T_m = 25 - 150 = -125°\text{C}$ $\Delta T_w = 25 - 20 = 5°\text{C}$

Step 3: Solve for $c_m$ . $m_m c_m \Delta T_m = -m_w c_w \Delta T_w$

Comparison with known metals:

Copper: 387 J/kg·°C
Steel: 470 J/kg·°C

Answer: Specific heat is 419 J/kg·°C

This is close to copper (387) or possibly a copper alloy.

Heat and Specific Heat

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🔥 Heat and Specific Heat

Heat vs. Temperature

Units of Heat

Specific Heat Capacity

📚 Practice Problems

1Problem 1easy

❓ Question:

Practice Lab

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📌 Related Topics in Thermodynamics

❓ Frequently Asked Questions

Common Specific Heats:

Sign Convention

Calorimetry

Typical Setup:

Heat Capacity

Methods of Heat Transfer

1. Conduction

2. Convection

3. Radiation

Problem-Solving Strategy

Common Mistakes

2Problem 2easy

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3Problem 3medium

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5Problem 5medium

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6Problem 6hard

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8Problem 8hard

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